How do you factor #y= 2x^3-2x^2-4x #?

1 Answer
Dec 26, 2015

Answer:

#y = 2x^3-2x^2-4x = 2x(x+1)(x-2)#

Explanation:

First note that all of the terms are divisible by #2x#, so separate that out as a factor:

#y = 2x^3-2x^2-4x = 2x(x^2-x-2)#

Then notice that if you substitute #x=-1# into #x^2-x-2#, the result is #0#, so #(x+1)# is a factor and the other factor must be #(x-2)# to get the #-2# constant term when multiplied:

#(x^2-x-2) = (x+1)(x-2)#

So we have:

#y = 2x^3-2x^2-4x = 2x(x+1)(x-2)#